Difference between revisions of "Group extension"

Latest revision as of 20:53, 27 May 2008

Let $F$ and $G$ be groups. An extension of $G$ by $F$ is a solution to the problem of finding a group $E$ that contains a normal subgroup $F'$ isomorphic to $F$ such that the quotient group $E/F'$ is isomorphic to $G$ .

More specifically, an extension $\mathcal{E}$ of $G$ by $F$ is a triple $(E,i,p)$ where $E$ is a group, $i$ is an injective group homomorphism of $F$ into $E$ , and $p$ is a surjective homomorphism of $E$ onto $G$ such that the kernel of $p$ is the image of $i$ . Often, the extension $\mathcal{E}$ is written as the diagram $\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G$ .

An extension is central if $i(F)$ lies in the center of $E$ ; this is only possible if $F$ is commutative. It is called trivial if $E = F \times G$ (the direct product of $F$ and $G$ ), $i$ is the canonical mapping of $F$ into $F\times G$ , and $p$ is the projection homomorphism onto $G$ .

Let $\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G$ and $\mathcal{E}' : F \stackrel{i'}{\to} E' \stackrel{p'}{\to} G$ be two extensions of $G$ by $F$ . A morphism of extensions from $\mathcal{E}$ to $\mathcal{E}'$ is a homomorphism $f: E \to E'$ such that $f \circ i = i'$ and $p' \circ f = p$ .

A retraction of an extension is a homomorphism $r: E\to F$ such that $r\circ i$ is the identity function on $F$ . Similarly, a section of an extension is a homomorphism $s : G \to E$ such that $p \circ s = \text{Id}_G$ .

Equivalence of Extensions

Theorem 1. Let $\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G$ and $\mathcal{E}' : F \stackrel{i'}{\to} E' \stackrel{p'}{\to} G$ be extensions of $G$ by $F$ . Let $u : E \to E'$ be a morphism of extensions. Then $u$ is an isomorphism of $E$ onto $E'$ . In other words, every extension morphism is an extension isomorphism.

Proof. Suppose $a,b$ are elements of $E$ such that $u(a)=u(b)$ . Then $p(a) = (p' \circ u)(a) = (p' \circ u)(b) = p(b) ,$ so $ab^{-1} \in \text{Ker}(p) = \text{Im}(i)$ . Let $g \in F$ be an element such that $i(g) = ab^{-1}$ . Then $e_{E'} = u(a)u(b)^{-1} = u(ab^{-1}) = (u \circ i)(g) = i'(g) .$ It follows that $g$ is the identity of $F$ , so $ab^{-1}$ is the identity of $F$ and $a=b$ .

Let $a'$ be in $E'$ ; since $p$ is surjective, there exists $a\in F$ such that $p'(a') = p(a) = p'(u(a))$ . Then $a' u(a)^{-1} \in \text{Ker}(p') = \text{Im}(i') = \text{Im}(u \circ i) \subseteq \text{Im}(u).$ Thus there exists $b \in F$ such that $u(b) = a'u(a)^{-1}$ ; then $u(ba) = a'$ . Thus $u$ is surjective. $\blacksquare$

Theorem 2. Let $\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G$ be an extension of $G$ by $F$ . Then the following are equivalent:

$\mathcal{E}$ is the trivial extension;
$\mathcal{E}$ admits a retraction;
$\mathcal{E}$ admits a section $s$ such that $s(G)$ lies in the centralizer of $i(F)$ .

Proof. If $\mathcal{E}$ is the trivial extension, then the projection onto $F$ and the canonical injection of $G$ into $E$ show that conditions 2 and 3 are satisfied. If $\mathcal{E}$ has a retraction $r$ , then the mapping $(r,p): E \to F \times G$ is an extension morphism, so $\mathcal{E}$ is isomorphic to the trivial extension. If (3) holds, then the mapping $(f,g) \mapsto i(f)s(g)$ is an extension morphism $F\times G \to \mathcal{E}$ , so again $\mathcal{E}$ is isomorphic to the trivial extension of $G$ by $F$ . $\blacksquare$

Note that an extension $F \stackrel{i}{\to} E \stackrel{p}{\to} G$ may be nontrivial, but $E$ may still be isomorphic to $F \times G$ .

@@ Line 1: / Line 1: @@
 Let <math>F</math> and <math>G</math> be [[group]]s.  An '''extension''' of <math>G</math> by <math>F</math> is a solution to the problem of finding a group <math>E</math> that contains a [[normal subgroup]] <math>F'</math> [[isomorphic]] to <math>F</math> such that the [[quotient group]] <math>E/F'</math> is isomorphic to <math>G</math>.
-More specifically, an extension <math>\mathcal{E}</math> of <math>F</math> by <math>G</math> is a triple <math>(E,i,p)</math> where <math>E</math> is a group, <math>i</math> is an [[injective]] group [[homomorphism]] of <math>F</math> into <math>E</math>, and <math>p</math> is a [[surjective]] homomorphism of <math>E</math> onto <math>G</math> such that the [[kernel]] of <math>p</math> is the image of <math>i</math>. Often, the extension <math>\mathcal{E}</math> is written as the diagram <math>\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G</math>.
+More specifically, an extension <math>\mathcal{E}</math> of <math>G</math> by <math>F</math> is a triple <math>(E,i,p)</math> where <math>E</math> is a group, <math>i</math> is an [[injective]] group [[homomorphism]] of <math>F</math> into <math>E</math>, and <math>p</math> is a [[surjective]] homomorphism of <math>E</math> onto <math>G</math> such that the [[kernel]] of <math>p</math> is the image of <math>i</math>. Often, the extension <math>\mathcal{E}</math> is written as the diagram <math>\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G</math>.
 An extension is '''central''' if <math>i(F)</math> lies in the [[center]] of <math>E</math>; this is only possible if <math>F</math> is [[abelian group | commutative]].  It is called ''trivial'' if <math>E = F \times G</math> (the direct product of <math>F</math> and <math>G</math>), <math>i</math> is the canonical mapping of <math>F</math> into <math>F\times G</math>, and <math>p</math> is the [[projection homomorphism]] onto <math>G</math>.
@@ Line 9: / Line 9: @@
 A ''retraction'' of an extension is a homomorphism <math>r: E\to F</math> such that <math>r\circ i</math> is the identity function on <math>F</math>.  Similarly, a ''section'' of an extension is a homomorphism <math>s : G \to E</math> such that <math>p \circ s = \text{Id}_G</math>.
-{{stub}}
+== Equivalence of Extensions ==
+'''Theorem 1.'''  Let <math>\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G</math> and <math>\mathcal{E}' : F \stackrel{i'}{\to} E' \stackrel{p'}{\to} G</math> be extensions of <math>G</math> by <math>F</math>.  Let <math>u : E \to E'</math> be a morphism of extensions.  Then <math>u</math> is an isomorphism of <math>E</math> onto <math>E'</math>.  In other words, every extension morphism is an extension isomorphism.
+''Proof.''  Suppose <math>a,b</math> are elements of <math>E</math> such that <math>u(a)=u(b)</math>.  Then
+<cmath> p(a) = (p' \circ u)(a) = (p' \circ u)(b) = p(b) , </cmath>
+so <math>ab^{-1} \in \text{Ker}(p) = \text{Im}(i)</math>.  Let <math>g \in F</math> be an element such that <math>i(g) = ab^{-1}</math>.  Then
+<cmath> e_{E'} = u(a)u(b)^{-1} = u(ab^{-1}) = (u \circ i)(g) = i'(g) . </cmath>
+It follows that <math>g</math> is the identity of <math>F</math>, so <math>ab^{-1}</math> is the identity of <math>F</math> and <math>a=b</math>.
+Let <math>a'</math> be in <math>E'</math>; since <math>p</math> is surjective, there exists <math>a\in F</math> such that <math>p'(a') = p(a) = p'(u(a))</math>.  Then
+<cmath> a' u(a)^{-1} \in \text{Ker}(p') = \text{Im}(i') = \text{Im}(u \circ i) \subseteq \text{Im}(u). </cmath>
+Thus there exists <math>b \in F</math> such that <math>u(b) = a'u(a)^{-1}</math>; then <math>u(ba) = a'</math>. Thus <math>u</math> is surjective.  <math>\blacksquare</math>
+'''Theorem 2.''' Let <math>\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G</math> be an extension of <math>G</math> by <math>F</math>.  Then the following are equivalent:
+# <math>\mathcal{E}</math> is the trivial extension;
+# <math>\mathcal{E}</math> admits a retraction;
+# <math>\mathcal{E}</math> admits a section <math>s</math> such that <math>s(G)</math> lies in the [[centralizer]] of <math>i(F)</math>.
+''Proof.''  If <math>\mathcal{E}</math> is the trivial extension, then the projection onto <math>F</math> and the canonical injection of <math>G</math> into <math>E</math> show that conditions 2 and 3 are satisfied.  If <math>\mathcal{E}</math> has a retraction <math>r</math>, then the mapping <math>(r,p): E \to F \times G</math> is an extension morphism, so <math>\mathcal{E}</math> is isomorphic to the trivial extension.  If (3) holds, then the mapping <math>(f,g) \mapsto i(f)s(g)</math> is an extension morphism <math>F\times G \to \mathcal{E}</math>, so again <math>\mathcal{E}</math> is isomorphic to the trivial extension of <math>G</math> by <math>F</math>.  <math>\blacksquare</math>
+Note that an extension <math>F \stackrel{i}{\to} E \stackrel{p}{\to} G</math> may be nontrivial, but <math>E</math> may still be isomorphic to <math>F \times G</math>.
+== See also ==
+* [[Semi-direct product]]
+* [[Simple group]]
+* [[Normal subgroup]]
+* [[Quotient group]]
 [[Category:Group theory]]

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Difference between revisions of "Group extension"

Latest revision as of 20:53, 27 May 2008

Equivalence of Extensions

See also